Solution :
Since A, B and C are the interior angles of a triangle ABC
\(\therefore\) A + B + C = 180
\(\implies\) \(A\over 2\) + \(B\over 2\) + \(C\over 2\) = 90
\(\implies\) \(B\over 2\) + \(C\over 2\) = 90 – \(A\over 2\)
\(\implies\) sin(\(B+C\over 2\)) = sin(90 – \(A\over 2\))
\(\implies\) sin\(B+C\over 2\) = cos\(A\over 2\).
